Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 3 Trigonometric Functions Ex 3.3 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.3
(ii) cosec-1(2)
Solution:
The principal value branch of cosec-1x is – {0}.
Let cosec-1(2) = ∝, where ≤ ∝ ≤ , ∝ ≠ 0
∴ cosec-1∝ = 2 = cosec
∴ ∝ = …[∵ – ≤ ≤ ]
∴ the principal value of cosec-1(2) is .
(iii) tan-1(-1)
Solution:
The principal value branch of tan-1x is
Let tan-1(-1) = ∝, where < ∝ <
∴ tan∝ = -1 = -tan
∴ tan∝ = tan …[∵ tan(-θ) = -tanθ]
∴ ∝ = – …[∵ – < < ]
∴ the principal value of tan-1(-1) is –.
(iv) tan-1(-)
Solution:
The principal value branch of tan-1x is .
Let tan-1(-) = ∝, where < ∝ <
∴ tan∝ = – = -tan
∴ tan∝ = tan …[∵ tan(-θ) = -tanθ]
∴ ∝ = – …[∵ – < < ]
∴ the principal value of tan-1(-) is –.
(v) sin-1
Solution:
The principal value branch of sin-1x is .
Let sin-1 = ∝, where < ∝ <
∴ sin∝ = = sin
∴ ∝ = …[∵ – ≤ ≤ ]
∴ the principal value of sin-1 is .
(vi) cos-1
Solution:
The principal value branch of cos-1x is (0, π).
Let cos-1 = ∝, where 0 ≤ ∝ ≤ π
∴ cos∝ = = -cos
∴ cos∝ = cos …[∵ cos(π – θ) = -cosθ)
∴ cos∝ = cos
∴ ∝ = …[∵ 0 ≤ ≤ π]
∴ the principal value of cos-1 is .
(ii) cos-1 + 2 sin-1
Solution:
(iii) tan-1 – sec-1(-2)
Solution:
∴ tan-1 – sec-1(-2)
= …[By (1) and (2)]
= –.
(iv) cosec-1( ) + cot-1()
Solution:
(ii) sin-1 + cos-1 = cos-1
Solution:
(iii) sin-1 + cos-1 = sin-1
Solution:
(iv) cos-1 + cos-1 =
Solution:
Let cos-1 = x
∴ cosx = , where 0 < x < ∴ sinx > 0
(v) tan-1 + tan-1 =
Solution:
LHS = tan-1 + tan-1
= tan-1
= tan-1 = tan-1(1)
= tan-1 =
= RHS.
(vi) 2 tan-1 = tan-1
Solution:
(vii) tan-1 = + θ if θ ∈
Solution:
(viii) tan-1, if θ ∈ (0, π)
Solution:
= …[∵ tan-1(tanθ) = θ]
= RHS.